Question

Find the QR factorization of the matrices$\left[\begin{array}{cc}1& 6\\ 1& 4\\ 1& 6\\ 1& 4\end{array}\right]$.

Short answer

The QR factorization of the matrix is $\left[\begin{array}{cc}1& 6\\ 1& 4\\ 1& 6\\ 1& 4\end{array}\right]=\left[\begin{array}{cc}1/2& 1/2\\ 1/2& -1/2\\ 1/2& 1/2\\ 1/2& -1/2\end{array}\right]\left[\begin{array}{cc}2& 10\\ 0& 2\end{array}\right]$.

Step-by-step solution

Determine column u→1 and entries r11 of R.

Consider the matrix $M=\left[\begin{array}{cc}1& 6\\ 1& 4\\ 1& 6\\ 1& 4\end{array}\right]\mathrm{where}{\overrightarrow{\mathrm{v}}}_1=\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]\mathrm{and}{\overrightarrow{\mathrm{v}}}_2=\left[\begin{array}{c}6\\ 4\\ 6\\ 4\end{array}\right]$

By the theorem of QR method, the value of ${\overrightarrow{u}}_1$and $r_{11}$is defined as follows.

$$\begin{gathered} r_{11}=||{\overrightarrow{v}}_1|| \\ {\overrightarrow{u}}_1=\frac{1}{r_{11}}{\overrightarrow{v}}_1 \end{gathered}$$

Simplify the equation $r_{11}=||{\overrightarrow{v}}_1||$as follows.

$$\begin{gathered} r_{11}=||{\overrightarrow{v}}_1|| \\ {r}_{11}=||\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]|| \\ {r}_{11}=\sqrt{1^2+1^2+1^2+1^2} \\ {r}_{11}=2 \end{gathered}$$

Substitute the values 2 for $r_{11}$and $\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]$for ${\overrightarrow{v}}_1$in the equation ${\overrightarrow{u}}_1=\frac{1}{r_{11}}{\overrightarrow{v}}_1$as follows.

role="math" localid="1659959656431" $$\begin{gathered} {\overrightarrow{u}}_1=\frac{1}{r_{11}}{\overrightarrow{v}}_1 \\ {\overrightarrow{u}}_1=\frac{1}{2}\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right] \\ {\overrightarrow{u}}_1=\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right] \end{gathered}$$

Therefore, the values${\overrightarrow{u}}_1=\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right]$ and $r_{11}=2$.

Determine column v2⊥ and entries r12 of R .

As ,$r_{12}={\overrightarrow{u}}_1.{\overrightarrow{v}}_2$ substitute the values $\left[\begin{array}{c}6\\ 4\\ 6\\ 4\end{array}\right]$for ${\overrightarrow{v}}_2$and $\left[\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}\right]$for ${\overrightarrow{u}}_1$in the equation $r_{11}={\overrightarrow{u}}_1.{\overrightarrow{v}}_2$as follows.

$$\begin{gathered} r_{12}={\overrightarrow{u}}_1.{\overrightarrow{v}}_2 \\ {r}_{12}=\left[\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}\right].\left[\begin{array}{c}6\\ 4\\ 6\\ 4\end{array}\right] \\ {r}_{12}=3+2+3+2 \\ {r}_{12}=10 \end{gathered}$$

Substitute the values $\left[\begin{array}{c}6\\ 4\\ 6\\ 4\end{array}\right]$for ${\overrightarrow{v}}_2$, 10 for $r_{12}$and $\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right]$for ${\overrightarrow{u}}_1$in the equation

${\overrightarrow{V}}_2^{\perp }=\overrightarrow{v_2}-r_{12}{\overrightarrow{u}}_1$as follows.

role="math" localid="1659960325676" $$\begin{gathered} {\overrightarrow{V}}_2^{\perp }=\overrightarrow{v_2}-r_{12}{\overrightarrow{u}}_1 \\ {\overrightarrow{V}}_2^{\perp }=\left[\begin{array}{c}6\\ 4\\ 6\\ 4\end{array}\right]-10\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right] \\ {\overrightarrow{V}}_2^{\perp }=\left[\begin{array}{c}6\\ 4\\ 6\\ 4\end{array}\right]-\left[\begin{array}{c}5\\ 5\\ 5\\ 5\end{array}\right] \\ {\overrightarrow{V}}_2^{\perp }=\left[\begin{array}{c}1\\ -1\\ 1\\ -1\end{array}\right] \end{gathered}$$

Therefore, the values${\overrightarrow{V}}_2^{\perp }=\left[\begin{array}{c}1\\ -1\\ 1\\ -1\end{array}\right]$ and role="math" localid="1659960392325" $r_{12}=10$.

Determine column u→2 and entries r22 of R.

The value of${\overrightarrow{u}}_2$and$r_{22}$is defined as follows.

$$\begin{gathered} r_{22}=||{\overrightarrow{v}}_2^{\perp }|| \\ {\overrightarrow{u}}_2=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp } \end{gathered}$$

Simplify the equation $r_{22}=||{\overrightarrow{v}}_2^{\perp }||$as follows.

$$\begin{gathered} r_{22}=||{\overrightarrow{v}}_2^{\perp }|| \\ {r}_{22}=||\left[\begin{array}{c}1\\ -1\\ 1\\ -1\end{array}\right]|| \\ {r}_{22}=\sqrt{{\left(1\right)}^2+{\left(-1\right)}^2+{\left(1\right)}^2+{\left(-1\right)}^2} \\ {r}_{22}=2 \end{gathered}$$

Substitute the values 2 for $r_{22}$and $\left[\begin{array}{c}1\\ -1\\ 1\\ -1\end{array}\right]$for ${\overrightarrow{v}}_2^{\perp }$in the equation ${\overrightarrow{u}}_2=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp }$as follows.

$$\begin{gathered} \overrightarrow{u}=\frac{1}{r_{22}}{\overrightarrow{v}}_2^{\perp } \\ \overrightarrow{u}=\frac{1}{2}\left[\begin{array}{c}1\\ -1\\ 1\\ -1\end{array}\right] \end{gathered}$$

$\overrightarrow{u_2}=\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right]$

The values role="math" localid="1659960881585" $\overrightarrow{{\mathrm{u}}_2}=\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right]$and $r_{22}=2$.

Therefore, the matrices $Q=\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\begin{array}{c}1/2\\ -1/2\\ 1/2\\ -1/2\end{array}\right]$and $R=\left[\begin{array}{cc}2& 10\\ 0& 2\end{array}\right]$.

Hence, the QR factorization of the matrix is $\left[\begin{array}{cc}1& 6\\ 1& 4\\ 1& 6\\ 1& 4\end{array}\right]=\left[\begin{array}{cc}1/2& 1/2\\ 1/2& -1/2\\ 1/2& 1/2\\ 1/2& -1/2\end{array}\right]\left[\begin{array}{cc}2& 10\\ 0& 2\end{array}\right]$.